Question: Let $a(x)=x^6-3x^5+x^4+x^2-3$, and $b(x)=x^4$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Explanation: Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's rewrite the fraction to cancel common factors: $ \begin{aligned} \dfrac{a(x)}{b(x)}=\dfrac{x^6-3x^5+x^4+x^2-3}{x^4}&=\dfrac{ {x^6}-3 {x^5}+ {x^4}}{ {x^4}}+\dfrac{x^2-3}{x^4}\\\\ &={x^2-3x+1}+\dfrac{{x^2-3}}{x^4}\\\\ &={q(x)} + \dfrac{{r(x)}}{b(x)}\end{aligned}$ Since the degree of ${x^2-3}$ is less than the degree of $x^4$, it follows that ${r(x)}={x^2-3}$, and ${q(x)}={x^2-3x+1}$. To conclude, $q(x)=x^2-3x+1$ $r(x)=x^2-3$ [Is there another way of doing this?]